AP sum formula — derivation and when to use nth term vs sum formula

Question

Find the sum of the first 20 terms of the AP: 3, 7, 11, 15, … Also, if the sum of first $n$ terms of an AP is $S_n = 3n^2 + 5n$ , find the 10th term.

Solution — Step by Step

Here $a = 3$ , $d = 7 - 3 = 4$ , $n = 20$ .

S_n = \frac{n}{2}[2a + (n-1)d] = \frac{20}{2}[2(3) + 19(4)] = 10[6 + 76] = 10 \times 82 = \mathbf{820}

The $n$ th term of an AP can be found from the sum formula: $a_n = S_n - S_{n-1}$

a_{10} = S_{10} - S_9 = [3(100) + 5(10)] - [3(81) + 5(9)]

= [300 + 50] - [243 + 45] = 350 - 288 = \mathbf{62}

Why This Works

graph TD
    A["AP Problem: What is asked?"] --> B["Find a specific term?"]
    A --> C["Find the sum of n terms?"]
    A --> D["Given S_n, find a term?"]
    B --> E["Use a_n = a + n-1 d"]
    C --> F["Use S_n = n/2 times 2a + n-1 d"]
    C --> G["Or S_n = n/2 times a + l if last term known"]
    D --> H["Use a_n = S_n - S_{n-1}"]

The sum formula comes from a beautiful trick by Gauss. Write the AP forward and backward:

$S_n = a + (a+d) + (a+2d) + ... + l$

$S_n = l + (l-d) + (l-2d) + ... + a$

Add term by term: $2S_n = n(a+l)$ , so $S_n = \frac{n(a+l)}{2}$ . Substituting $l = a + (n-1)d$ gives the standard formula.

The $a_n = S_n - S_{n-1}$ trick works because $S_n$ is the sum of first $n$ terms and $S_{n-1}$ is the sum of first $n-1$ terms. The difference must be the $n$ th term alone.

In JEE Main, if $S_n$ is given as a quadratic in $n$ (like $S_n = 3n^2 + 5n$ ), the sequence is always an AP. The common difference is twice the coefficient of $n^2$ . Here $d = 2 \times 3 = 6$ . This shortcut saves time.

Alternative Method

When you know the first term $a$ and last term $l$ , use the simpler formula: $S_n = \frac{n}{2}(a + l)$ . This is faster when the last term is directly given or easy to compute.

For the first part: $a_{20} = 3 + 19(4) = 79$ . So $S_{20} = \frac{20}{2}(3 + 79) = 10 \times 82 = 820$ . Same answer, less arithmetic.

Common Mistake

Using $n$ instead of $n-1$ in the nth term formula. The formula is $a_n = a + (n-1)d$ , not $a + nd$ . The first term is $a_1 = a + 0 \cdot d = a$ , not $a + d$ . This off-by-one error is the most common mistake in AP problems. When in doubt, verify: $a_1 = a + (1-1)d = a$ ✓, $a_2 = a + (2-1)d = a + d$ ✓.

$n$ th term: $a_n = a + (n-1)d$

Sum of $n$ terms: $S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}(a + l)$

$n$ th term from sum: $a_n = S_n - S_{n-1}$ (for $n \geq 2$ )

If $S_n = An^2 + Bn$ , then $d = 2A$ and $a = A + B$

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next